Point by point, the transformation $T(x, y) = (4x - y, 3x -2y)$ sends the line $x + 2y = 6$ onto an image line. What is the slope of the image?

Question

The question is as follows:

Point by point, the transformation $T(x, y) = (4x - y, 3x -2y)$ sends the line $x + 2y = 6$ onto an image line. What is the slope of the image?

I don't understand what this question basically. How is it that the transformation is able to send a line into an "image line" (a term that I am not even familiarized with)? Any help will be greatly appreciated.

Answer 1

Think of $T$ here as sending all points $(x,y)$ satisfying $x+2y=6$ to $(x',y')=(4x-y,3x-2y)$. Since there is an equation between $x$ and $y$, there must also be one between $x'$ and $y'$, and this latter equation will be the image line. $$x'=4x-y$$ $$y'=3x-2y$$ $$-2x'=2y-8x$$ $$y'-2x'=-5x\qquad x=(2x'-y')/5\tag1$$ $$y'=3/5(2x'-y')-2y$$ $$2y=3/5(2x'-y')-y'=6/5x'-8/5y'$$ $$y=(3x'-4y')/5\tag2$$ Then substituting $(1)$ and $(2)$ into $x+2y=6$: $$(2x'-y')/5+2(3x'-4y')/5=6$$ $$8x'-9y'=30$$ $$y'=(8x'-30)/9$$ This is the image line, and its slope is $\frac89$.

Answer 2

assume $h=4x-y$ and $k=3x-2y$ now take the image of the point $(h,k)$ with respect to the line $4x+2y=6$ you will get a point now you can get the slope easily

Answer 3

$T:\mathbb{R^2} \rightarrow \mathbb{R^2}$.

$T(x,y) \mapsto (x',y')$,

$x'=4x-y$; $y'=3x-2y.$

Solving for $x,y$ in terms of $x',y'$.

$2x'-y'=5x$; and $ 3x' -4y' =5y$.

The line $x+2y = 6$ is transformed:

$5x +10y =30 \rightarrow$

$(2x'-y' ) + 2 (3x'-4y') =30.$

$8x' - 9y' = 30.$

Slope of the image line: $m' = 8/9.$